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Tag 0CLQ

91.39. Valuative criterion for second diagonal

The converse statement has already been proved in Lemma 91.38.2. The criterion itself is the following.

Lemma 91.39.1. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. If $\Delta_f$ is quasi-separated and if for every diagram (91.38.1.1) and choice of $\gamma$ as in Definition 91.38.1 the category of dotted arrows is a setoid, then $\Delta_f$ is separated.

Proof. We are going to write out a detailed proof, but we strongly urge the reader to find their own proof, inspired by reading the argument given in the proof of Lemma 91.38.2.

Assume $\Delta_f$ is quasi-separated and for every diagram (91.38.1.1) and choice of $\gamma$ as in Definition 91.38.1 the category of dotted arrows is a setoid. By Lemma 91.6.1 it suffices to show that $e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ is a closed immersion. By Lemma 91.6.4 it in fact suffices to show that $e = \Delta_{f, 2}$ is universally closed. Either of these lemmas tells us that $e = \Delta_{f, 2}$ is quasi-compact by our assumption that $\Delta_f$ is quasi-separated.

In this paragraph we will show that $e$ satisfies the existence part of the valuative criterion. Consider a $2$-commutative solid diagram $$ \xymatrix{ \mathop{\rm Spec}(K) \ar[r]_x \ar[d]_j & \mathcal{X} \ar[d]^e \\ \mathop{\rm Spec}(A) \ar[r]^{(a, \theta)} & \mathcal{I}_{\mathcal{X}/\mathcal{Y}} } $$ and let $\alpha : (a, \theta) \circ j \to e \circ x$ be any $2$-morphism witnessing the $2$-commutativity of the diagram (we use $\alpha$ instead of the letter $\gamma$ used in Definition 91.38.1). Note that $f \circ \theta = \text{id}$; we will use this below. Observe that $e \circ x = (x, \text{id}_x)$ and $(a, \theta) \circ j = (a \circ j, \theta \star \text{id}_j)$. Thus we see that $\alpha$ is a $2$-arrow $\alpha : a \circ j \to x$ compatible with $\theta \star \text{id}_j$ and $\text{id}_x$. Set $y = f \circ x$ and $\beta = \text{id}_{f \circ a}$. Reading the arguments given in the proof of Lemma 91.38.2 backwards, we see that $\theta$ is an automorphism of the dotted arrow $(a, \alpha, \beta)$ with $$ \gamma : y \circ j \to f \circ x \quad\text{equal to}\quad \text{id}_f \star \alpha : f \circ a \circ j \to f \circ x $$ On the other hand, $\text{id}_a$ is an automorphism too, hence we conclude $\theta = \text{id}_a$ from the assumption on $f$. Then we can take as dotted arrow for the displayed diagram above the morphism $a : \mathop{\rm Spec}(A) \to \mathcal{X}$ with $2$-morphisms $(a, \text{id}_a) \circ j \to (x, \text{id}_x)$ given by $\alpha$ and $(a, \theta) \to e \circ a$ given by $\text{id}_a$.

By Lemma 91.38.11 any base change of $e$ satisfies the existence part of the valuative criterion. Since $e$ is representable by algebraic spaces, it suffices to show that $e$ is universally closed after a base change by a morphism $I \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$ which is surjective and smooth and with $I$ an algebraic space (see Properties of Stacks, Section 90.3). This base change $e' : X' \to I'$ is a quasi-compact morphism of algebraic spaces which satisfies the existence part of the valuative criterion and hence is universally closed by Morphisms of Spaces, Lemma 58.41.1. $\square$

    The code snippet corresponding to this tag is a part of the file stacks-morphisms.tex and is located in lines 8966–9058 (see updates for more information).

    \section{Valuative criterion for second diagonal}
    \label{section-valuative-second}
    
    \noindent
    The converse statement has already been proved in
    Lemma \ref{lemma-cat-dotted-arrows}.
    The criterion itself is the following.
    
    \begin{lemma}
    \label{lemma-setoids-and-diagonal}
    Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks.
    If $\Delta_f$ is quasi-separated and if for every diagram
    (\ref{equation-diagram}) and choice of $\gamma$ as in
    Definition \ref{definition-fill-in-diagram}
    the category of dotted arrows
    is a setoid, then $\Delta_f$ is separated.
    \end{lemma}
    
    \begin{proof}
    We are going to write out a detailed proof, but we strongly urge the
    reader to find their own proof, inspired by reading the argument
    given in the proof of Lemma \ref{lemma-cat-dotted-arrows}.
    
    \medskip\noindent
    Assume $\Delta_f$ is quasi-separated and for every diagram
    (\ref{equation-diagram}) and choice of $\gamma$ as in
    Definition \ref{definition-fill-in-diagram}
    the category of dotted arrows is a setoid.
    By Lemma \ref{lemma-diagonal-diagonal} it suffices to show that
    $e : \mathcal{X} \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$
    is a closed immersion. By
    Lemma \ref{lemma-first-diagonal-separated-second-diagonal-closed}
    it in fact suffices to show that $e = \Delta_{f, 2}$ is
    universally closed.
    Either of these lemmas tells us that $e = \Delta_{f, 2}$ is quasi-compact
    by our assumption that $\Delta_f$ is quasi-separated.
    
    \medskip\noindent
    In this paragraph we will show that $e$ satisfies the existence
    part of the valuative criterion. Consider a $2$-commutative solid diagram
    $$
    \xymatrix{
    \Spec(K) \ar[r]_x \ar[d]_j & \mathcal{X} \ar[d]^e \\
    \Spec(A) \ar[r]^{(a, \theta)} & \mathcal{I}_{\mathcal{X}/\mathcal{Y}}
    }
    $$
    and let $\alpha : (a, \theta) \circ j \to e \circ x$ be any $2$-morphism
    witnessing the $2$-commutativity of the diagram (we use $\alpha$ instead
    of the letter $\gamma$ used in Definition \ref{definition-fill-in-diagram}).
    Note that $f \circ \theta = \text{id}$; we will use this below.
    Observe that $e \circ x = (x, \text{id}_x)$ and
    $(a, \theta) \circ j = (a \circ j, \theta \star \text{id}_j)$.
    Thus we see that $\alpha$ is a $2$-arrow $\alpha : a \circ j \to x$
    compatible with $\theta \star \text{id}_j$ and $\text{id}_x$.
    Set $y = f \circ x$ and $\beta = \text{id}_{f \circ a}$.
    Reading the arguments given in the proof of
    Lemma \ref{lemma-cat-dotted-arrows}
    backwards, we see that $\theta$ is an automorphism of the
    dotted arrow $(a, \alpha, \beta)$ with
    $$
    \gamma : y \circ j \to f \circ x
    \quad\text{equal to}\quad
    \text{id}_f \star \alpha : f \circ a \circ j \to f \circ x
    $$
    On the other hand, $\text{id}_a$ is an automorphism too, hence
    we conclude $\theta = \text{id}_a$ from the assumption on $f$.
    Then we can take as dotted arrow for the displayed diagram above
    the morphism $a : \Spec(A) \to \mathcal{X}$ with $2$-morphisms
    $(a, \text{id}_a) \circ j \to (x, \text{id}_x)$ given by $\alpha$
    and $(a, \theta) \to e \circ a$ given by $\text{id}_a$.
    
    \medskip\noindent
    By Lemma \ref{lemma-base-change-existence} any base change of $e$
    satisfies the existence part of the valuative criterion.
    Since $e$ is representable by algebraic spaces, it suffices to
    show that $e$ is universally closed after a base change
    by a morphism $I \to \mathcal{I}_{\mathcal{X}/\mathcal{Y}}$
    which is surjective and smooth and with $I$ an algebraic space
    (see Properties of Stacks, Section
    \ref{stacks-properties-section-properties-morphisms}).
    This base change $e' : X' \to I'$ is a quasi-compact
    morphism of algebraic spaces which
    satisfies the existence part of the valuative criterion
    and hence is universally closed by
    Morphisms of Spaces, Lemma
    \ref{spaces-morphisms-lemma-quasi-compact-existence-universally-closed}.
    \end{proof}

    Comments (2)

    Comment #2100 by Matthew Emerton on June 22, 2016 a 7:44 pm UTC

    Does always a setoid'' meanfor any choice of $\gamma$''? If so, it would be good to make this explicit, since the fact that we have to allow this for every choice of $\gamma$ rather than just our favourite choice seems to be a slightly subtle point.

    Comment #2144 by Johan (site) on July 22, 2016 a 2:54 pm UTC

    OK, yes you are right. Fixed here.

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